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One more simple proof of the Craig–Sakamoto theorem
Linear Algebra and its Applications 431 (2009) 1616–1619
Contents lists available at ScienceDirect
Linear Algebra and its Applications journal homepage: www.elsevier.com/locate/laa
One more simple proof of the Craig–Sakamoto theorem Hervé Carrieu a , Patrice Lassère b,∗ a
Lycée Camille Guérin, 86000 Poitiers, France
b
Institut de Mathématiques, UMR CNRS 5580, Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France
A R T I C L E
I N F O
Article history: Received 26 May 2008 Accepted 25 May 2009 Available online 9 July 2009
A B S T R A C T
We give one more elementary proof of the Craig–Sakamoto’s theorem: given A, B ∈ Sn (R) such that det(In − xA − yB) = det(In − xA) det(In − yB), ∀ x, y ∈ R; then AB = 0. © 2009 Elsevier Inc. All rights reserved.
Submitted by C.K. Li AMS classification: 15A15 15A18 15A24 15A48
Keywords: Craig–Sakamoto’s theorem Linear and bilinear algebra
1. Introduction The Craig–Sakamoto theorem asserts that Given A, B ∈ Sn (R) then det(In − xA − yB) = det(In − xA) det(In − yB), ∀ x, y ∈ R if and only if AB = 0. For an historical viewpoint of this result coming from statistical-probabilities, the interested reader can look at [1,2,5,6]; since his first statement it has inspired many proofs (see [3,4,8–10], and the references listed in the previous papers). The purpose of this note is to give one more new (let us hope…) proof of this theorem using the “elementary” machinery of linear and bilinear algebra.
∗ Corresponding author.
E-mail addresses: [email protected] (H. Carrieu), [email protected] (P. Lasse` re).
0024-3795/$ - see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2009.05.031
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1617
2. The proof The key of the proof is the following: Property (). Let A, B det(In
∈ Sn (R) \ {OMn (R) } such that
− xA − yB) = det(In − xA) det(In − yB), ∀x, y ∈ R.
Then B (or A, the same by symmetry) admits a nonzero eigenvalue λ such that ker(A) ∩ ker In − λ−1 B = / {0}. First, we are going to show how this property implies the Craig–Sakamoto’s theorem. 2.1. Proof of the Craig–Sakamoto’s theorem We do it by induction on the size n of the matrices. For n = 1, the Craig–Sakamoto’s theorem is clear, so let n 2 and suppose it true up to rank n − 1; let A, B ∈ Sn (R) \ {OMn (R) } (we exclude the trivial case where one of the two matrices is zero) such that det(In
− xA − yB) = det(In − xA) det(In − yB), ∀x, y ∈ R.
We have to prove that AB = 0. Because of property () there exists a nonzero eigenvalue of B, say λ, such that ker(A) ∩ ker(I − / {0}; so choose eλ ∈ ker(A) ∩ ker(I − λB) \ {0} and consider an orthogonal basis B of Rn with λB) = first term eλ . Because of the choice of eλ , the symmetrics matrices of A and B in the basis B have 0 0 λ 0 the respective shapes 0 A and 0 B where the matrices A and B belong to Sn−1 (R). An elementary computation gives
so
det(In
− xA − yB) = (1 − λy) det In−1 − xA − yB ,
det(In
− xA) det(In − yB) = (1 − λy) det In−1 − xA det In−1 − yB ,
det In−1
− xA − yB = det In−1 − xA det In−1 − yB , ∀ x, y ∈ R.
Then, by the induction hypothesis A B = 0, and we have 0 λ 0 0 0 −1 0 AB = P −1 P P = P 0 A 0 B 0 A B and we are done.
=0
2.2. Proof of property () For it, we first need two lemmas. Lemma 1. Let U = ((uij )) a symmetric positive semi-definite matrix; if a diagonal coefficient uii (1 i n) is equal to zero, then uij = uji = 0 for all 1 j n. Proof of the lemma 1. This is classical and elementary (see [7, problem 20.1]) but for this demonstration to be self-contained we include the proof: let U such a matrix with uii = 0 and, by contradiction, suppose that there exists a coefficient uji = / 0. Let Xt = (xk )n1 the vector where xj = 1, xi = tuji , t ∈ R and where the other components are zero, then t Xt UXt = ujj + 2tu2ji change sign when t runs through R which is impossible.
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1618
∈ Sn (R). Suppose U 0 and ∀ t ∈ R : det(U − tV ) = 0.
Lemma 2. Let U, V Then
ker(V ) ∩ ker(U )
= / {0}.
Remark. Notethat thislemma is obviously false without some symmetry hypothesis; for example, In−1 0 0 1 consider U = 0 0 , V = O 0 . n−1
Proof of the lemma 2. V is diagonalisable in an orthonormal basis: D 0 ∃P ∈ On (R) : PV t P = Diag(λ1 , . . . , λr , 0, . . . , 0) = r , 0 0 (note that r PU t P
< n because det(U ) = 0) then we have: − tPV t P =
U1 U2
t
U2 U3
−t
Dr 0
0 0
=
U1
− tDr
t
U2
U2 U3
λi = / 0
.
And in fact, it is possible to choose P (at the expense of changing the n − r lasts basis vectors) to have also U3 diagonal: precisely, let Q ∈ On−r so that QU3 t Q is diagonal, then, in the new basis associated I 0 t t to the orthogonal matrix 0r Q our matrix PU P − tPV P is U1 − tDr U2 t U2 QU3 t Q and is in the required shape. So let us consider such a P, because of the hypothesis, the polynomial
R t → det(PU t P − tD) = 0 is nul; the coefficient of t r being (up to a sign) λ1 . . . λr det(U3 ), we have det(U3 ) = 0. Now, det(U3 ) = 0 and U3 diagonal implies that the positive symmetric matrix PU t P admits a diagonal element equal to zero, say the i-th (i ∈ {r + 1, . . . , n}): then by lemma 1 the i-th column in PU t P is also null, e.g. PU t Pei = 0; but, because i ∈ {r + 1, . . . , n} we have also Dr 0 PV t Pei = e = 0. 0 0 i Consequently U t Pei
= V t Pei = 0, and finally t Pei ∈ ker(U ) ∩ ker(V ). What we had to prove.
Now the proof of the property () is easy: Proof of property (). Let λ be a non zero eigenvalue of B, we have det In − λ−1 B − xA = 0, ∀ x ∈ R and we will be in the case of the lemma 2 with V = A and U = In − λ−1 B who will be positive semi-definite if we choose λ as the greatest nonzero eigenvalue of B. Acknowledgments It is our pleasure to thank our colleague Jean Baptiste Hiriart-Urruty (University of Toulouse) who first told us about the Craig–Sakamoto’s theorem. References [1] M.F. Driscoll,W.R. Gundberg, A history of the development of Craig’s theorem, Am. Stat. 40 (1986) 65–70. [2] J.B. Hiriart-Urruty, Le théorème déterminental en bref, RMS 119 (2009) 15–22.
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1619
[3] Chi-Kwong Li, A simple proof of the Craig–Sakamoto’s theorem, Linear Algebra Appl., 321 (2000) 281–283. [4] M. Matsuura, On the Craig–Sakamoto’s theorem and Olkin’s determinantal result, Linear Algebra Appl. 364 (2003) 321–323. [5] J. Ogawa, A history of the development of Craig–Sakamoto’s theorem from Japanese standpoint, Proc. Ann. Inst. Stat. Math. 41 (1993) 47–59. [6] J. Ogawa, I. Olkin, A tale of two countries: The Craig–Sakamoto–Matusita theorem, J. Stat. Plann. Inference, in press. [7] V.V. Prasolov, Problems and Theorems in Linear Algebra, Translation of Mathematical Monographs, vol. 134, AMS, 1994. [8] T. Ogosawara, M. Takahashi, Independence of quadratics quantities in a normal system, J. Sci. Hiroshima Univ. 15 (1951) 1–9. [9] I. Olkin, A determinantal proof of the Craig–Sakamoto’s theorem, Linear Algebra Appl. 264 (1997) 217–223. [10] O. Taussky, On a matrix theorem of A.T. Craig and H. Hotelling, Indagationes Math. 20 (1958) 139–141.
Contents lists available at ScienceDirect
Linear Algebra and its Applications journal homepage: www.elsevier.com/locate/laa
One more simple proof of the Craig–Sakamoto theorem Hervé Carrieu a , Patrice Lassère b,∗ a
Lycée Camille Guérin, 86000 Poitiers, France
b
Institut de Mathématiques, UMR CNRS 5580, Université Paul Sabatier, 118 route de Narbonne, 31062 Toulouse, France
A R T I C L E
I N F O
Article history: Received 26 May 2008 Accepted 25 May 2009 Available online 9 July 2009
A B S T R A C T
We give one more elementary proof of the Craig–Sakamoto’s theorem: given A, B ∈ Sn (R) such that det(In − xA − yB) = det(In − xA) det(In − yB), ∀ x, y ∈ R; then AB = 0. © 2009 Elsevier Inc. All rights reserved.
Submitted by C.K. Li AMS classification: 15A15 15A18 15A24 15A48
Keywords: Craig–Sakamoto’s theorem Linear and bilinear algebra
1. Introduction The Craig–Sakamoto theorem asserts that Given A, B ∈ Sn (R) then det(In − xA − yB) = det(In − xA) det(In − yB), ∀ x, y ∈ R if and only if AB = 0. For an historical viewpoint of this result coming from statistical-probabilities, the interested reader can look at [1,2,5,6]; since his first statement it has inspired many proofs (see [3,4,8–10], and the references listed in the previous papers). The purpose of this note is to give one more new (let us hope…) proof of this theorem using the “elementary” machinery of linear and bilinear algebra.
∗ Corresponding author.
E-mail addresses: [email protected] (H. Carrieu), [email protected] (P. Lasse` re).
0024-3795/$ - see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2009.05.031
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1617
2. The proof The key of the proof is the following: Property (). Let A, B det(In
∈ Sn (R) \ {OMn (R) } such that
− xA − yB) = det(In − xA) det(In − yB), ∀x, y ∈ R.
Then B (or A, the same by symmetry) admits a nonzero eigenvalue λ such that ker(A) ∩ ker In − λ−1 B = / {0}. First, we are going to show how this property implies the Craig–Sakamoto’s theorem. 2.1. Proof of the Craig–Sakamoto’s theorem We do it by induction on the size n of the matrices. For n = 1, the Craig–Sakamoto’s theorem is clear, so let n 2 and suppose it true up to rank n − 1; let A, B ∈ Sn (R) \ {OMn (R) } (we exclude the trivial case where one of the two matrices is zero) such that det(In
− xA − yB) = det(In − xA) det(In − yB), ∀x, y ∈ R.
We have to prove that AB = 0. Because of property () there exists a nonzero eigenvalue of B, say λ, such that ker(A) ∩ ker(I − / {0}; so choose eλ ∈ ker(A) ∩ ker(I − λB) \ {0} and consider an orthogonal basis B of Rn with λB) = first term eλ . Because of the choice of eλ , the symmetrics matrices of A and B in the basis B have 0 0 λ 0 the respective shapes 0 A and 0 B where the matrices A and B belong to Sn−1 (R). An elementary computation gives
so
det(In
− xA − yB) = (1 − λy) det In−1 − xA − yB ,
det(In
− xA) det(In − yB) = (1 − λy) det In−1 − xA det In−1 − yB ,
det In−1
− xA − yB = det In−1 − xA det In−1 − yB , ∀ x, y ∈ R.
Then, by the induction hypothesis A B = 0, and we have 0 λ 0 0 0 −1 0 AB = P −1 P P = P 0 A 0 B 0 A B and we are done.
=0
2.2. Proof of property () For it, we first need two lemmas. Lemma 1. Let U = ((uij )) a symmetric positive semi-definite matrix; if a diagonal coefficient uii (1 i n) is equal to zero, then uij = uji = 0 for all 1 j n. Proof of the lemma 1. This is classical and elementary (see [7, problem 20.1]) but for this demonstration to be self-contained we include the proof: let U such a matrix with uii = 0 and, by contradiction, suppose that there exists a coefficient uji = / 0. Let Xt = (xk )n1 the vector where xj = 1, xi = tuji , t ∈ R and where the other components are zero, then t Xt UXt = ujj + 2tu2ji change sign when t runs through R which is impossible.
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1618
∈ Sn (R). Suppose U 0 and ∀ t ∈ R : det(U − tV ) = 0.
Lemma 2. Let U, V Then
ker(V ) ∩ ker(U )
= / {0}.
Remark. Notethat thislemma is obviously false without some symmetry hypothesis; for example, In−1 0 0 1 consider U = 0 0 , V = O 0 . n−1
Proof of the lemma 2. V is diagonalisable in an orthonormal basis: D 0 ∃P ∈ On (R) : PV t P = Diag(λ1 , . . . , λr , 0, . . . , 0) = r , 0 0 (note that r PU t P
< n because det(U ) = 0) then we have: − tPV t P =
U1 U2
t
U2 U3
−t
Dr 0
0 0
=
U1
− tDr
t
U2
U2 U3
λi = / 0
.
And in fact, it is possible to choose P (at the expense of changing the n − r lasts basis vectors) to have also U3 diagonal: precisely, let Q ∈ On−r so that QU3 t Q is diagonal, then, in the new basis associated I 0 t t to the orthogonal matrix 0r Q our matrix PU P − tPV P is U1 − tDr U2 t U2 QU3 t Q and is in the required shape. So let us consider such a P, because of the hypothesis, the polynomial
R t → det(PU t P − tD) = 0 is nul; the coefficient of t r being (up to a sign) λ1 . . . λr det(U3 ), we have det(U3 ) = 0. Now, det(U3 ) = 0 and U3 diagonal implies that the positive symmetric matrix PU t P admits a diagonal element equal to zero, say the i-th (i ∈ {r + 1, . . . , n}): then by lemma 1 the i-th column in PU t P is also null, e.g. PU t Pei = 0; but, because i ∈ {r + 1, . . . , n} we have also Dr 0 PV t Pei = e = 0. 0 0 i Consequently U t Pei
= V t Pei = 0, and finally t Pei ∈ ker(U ) ∩ ker(V ). What we had to prove.
Now the proof of the property () is easy: Proof of property (). Let λ be a non zero eigenvalue of B, we have det In − λ−1 B − xA = 0, ∀ x ∈ R and we will be in the case of the lemma 2 with V = A and U = In − λ−1 B who will be positive semi-definite if we choose λ as the greatest nonzero eigenvalue of B. Acknowledgments It is our pleasure to thank our colleague Jean Baptiste Hiriart-Urruty (University of Toulouse) who first told us about the Craig–Sakamoto’s theorem. References [1] M.F. Driscoll,W.R. Gundberg, A history of the development of Craig’s theorem, Am. Stat. 40 (1986) 65–70. [2] J.B. Hiriart-Urruty, Le théorème déterminental en bref, RMS 119 (2009) 15–22.
H. Carrieu, P. Lasse` re / Linear Algebra and its Applications 431 (2009) 1616–1619
1619
[3] Chi-Kwong Li, A simple proof of the Craig–Sakamoto’s theorem, Linear Algebra Appl., 321 (2000) 281–283. [4] M. Matsuura, On the Craig–Sakamoto’s theorem and Olkin’s determinantal result, Linear Algebra Appl. 364 (2003) 321–323. [5] J. Ogawa, A history of the development of Craig–Sakamoto’s theorem from Japanese standpoint, Proc. Ann. Inst. Stat. Math. 41 (1993) 47–59. [6] J. Ogawa, I. Olkin, A tale of two countries: The Craig–Sakamoto–Matusita theorem, J. Stat. Plann. Inference, in press. [7] V.V. Prasolov, Problems and Theorems in Linear Algebra, Translation of Mathematical Monographs, vol. 134, AMS, 1994. [8] T. Ogosawara, M. Takahashi, Independence of quadratics quantities in a normal system, J. Sci. Hiroshima Univ. 15 (1951) 1–9. [9] I. Olkin, A determinantal proof of the Craig–Sakamoto’s theorem, Linear Algebra Appl. 264 (1997) 217–223. [10] O. Taussky, On a matrix theorem of A.T. Craig and H. Hotelling, Indagationes Math. 20 (1958) 139–141.